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## 物理代写|分析力学代写Analytical Mechanics代考|Lagrange’s Equations

Equations (1.83) take a particularly concise and elegant form whenever the applied forces $\mathbf{F}_i$ can be derived from a scalar potential $V\left(\mathbf{r}_1, \ldots, \mathbf{r}_N, t\right)$. In this case,
$$\mathbf{F}_i=-\nabla_i V=-\left(\frac{\partial V}{\partial x_i} \hat{\mathbf{x}}+\frac{\partial V}{\partial y_i} \hat{\mathbf{y}}+\frac{\partial V}{\partial z_i} \hat{\mathbf{z}}\right)$$

and the generalised forces are written as
$$Q_k=\sum_{i=1}^N \mathbf{F}i \cdot \frac{\partial \mathbf{r}_i}{\partial q_k}=-\sum{i=1}^N\left(\frac{\partial V}{\partial x_i} \frac{\partial x_i}{\partial q_k}+\frac{\partial V}{\partial y_i} \frac{\partial y_i}{\partial q_k}+\frac{\partial V}{\partial z_i} \frac{\partial z_i}{\partial q_k}\right)=-\frac{\partial V}{\partial q_k}$$
where the chain rule of differentiation has been used. By means of Eqs. (1.71) the potential $V$ is expressed as a function of the $q$ s alone, being independent of the generalised velocities. Inserting (1.96) into (1.83) there results
$$\frac{d}{d t}\left(\frac{\partial T}{\partial \dot{q}_k}\right)-\frac{\partial}{\partial q_k}(T-V)=0 .$$
Given that $\partial V / \partial \dot{q}_k=0$, these last equations are equivalent to
$$\frac{d}{d t}\left[\frac{\partial}{\partial \dot{q}_k}(T-V)\right]-\frac{\partial}{\partial q_k}(T-V)=0 .$$
Defining the Lagrangian function or, simply, Lagrangian $L$ by
$$L=T-V,$$
the equations of motion for the system assume the form
$$\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_k}\right)-\frac{\partial L}{\partial q_k}=0, \quad k=1, \ldots, n$$

## 物理代写|分析力学代写Analytical Mechanics代考|Invariance of Lagrange’s Equations

Although the invariance of Lagrange’s equations under a general coordinate transformation is obvious from the above derivation, a direct proof is instructive. If $Q_1, \ldots, Q_n$ are new generalised coordinates which are differentiable functions of the original coordinates $q_1, \ldots, q_n$, we have ${ }^{15}$
$$Q_k=G_k\left(q_1, \ldots, q_n, t\right), \quad k=1, \ldots, n$$
and, conversely,
$$q_k=g_k\left(Q_1, \ldots, Q_n, t\right), \quad k=1, \ldots, n .$$
Invertibility requires the following condition on the Jacobian of the transformation:
$$\frac{\partial\left(q_1, \ldots, q_n\right)}{\partial\left(Q_1, \ldots, Q_n\right)} \equiv \operatorname{det}\left(\frac{\partial q_k}{\partial Q_l}\right)=\left(\frac{\partial\left(Q_1, \ldots, Q_n\right)}{\partial\left(q_1, \ldots, q_n\right)}\right)^{-1} \neq 0 .$$
The coordinate change (1.101) is called a point transformation because it maps points from the configuration space described by the $q s$ into points of the configuration space described by the $Q$ s. In mathematical terminology, a bijective differentiable application $G$ whose inverse $g=G^{-1}$ is also differentiable is called a diffeomorphism, and the configuration space of the $Q \mathrm{~s}$ is said to be diffeomorphic to the configuration space of the $q \mathrm{~s}$.
Taking the time derivative of Eq. (1.102), we find
$$\dot{q}k=\sum{l=1}^n \frac{\partial q_k}{\partial Q_l} \dot{Q}_l+\frac{\partial \dot{q}_k}{\partial t},$$
whence
$$\frac{\partial \dot{q}_k}{\partial \dot{Q}_l}=\frac{\partial q_k}{\partial Q_l}$$
The transformed Lagrangian $\bar{L}(Q, \dot{Q}, t)$ is just the original Lagrangian $L(q, \dot{q}, t)$ expressed in terms of $(Q, \dot{Q}, t)$
$$\bar{L}(Q, \dot{Q}, t)=L(q(Q, t), \dot{q}(Q, \dot{Q}, t), t)$$

# 分析力学代考

## 物理代写|分析力学代写分析力学代考|拉格朗日方程

$$\mathbf{F}_i=-\nabla_i V=-\left(\frac{\partial V}{\partial x_i} \hat{\mathbf{x}}+\frac{\partial V}{\partial y_i} \hat{\mathbf{y}}+\frac{\partial V}{\partial z_i} \hat{\mathbf{z}}\right)$$

$$Q_k=\sum_{i=1}^N \mathbf{F}i \cdot \frac{\partial \mathbf{r}_i}{\partial q_k}=-\sum{i=1}^N\left(\frac{\partial V}{\partial x_i} \frac{\partial x_i}{\partial q_k}+\frac{\partial V}{\partial y_i} \frac{\partial y_i}{\partial q_k}+\frac{\partial V}{\partial z_i} \frac{\partial z_i}{\partial q_k}\right)=-\frac{\partial V}{\partial q_k}$$
，其中使用了微分的链式法则。通过方程式。(1.71)势$V$仅表示为$q$ s的函数，与广义速度无关。将(1.96)代入(1.83)结果
$$\frac{d}{d t}\left(\frac{\partial T}{\partial \dot{q}_k}\right)-\frac{\partial}{\partial q_k}(T-V)=0 .$$

$$\frac{d}{d t}\left[\frac{\partial}{\partial \dot{q}_k}(T-V)\right]-\frac{\partial}{\partial q_k}(T-V)=0 .$$

$$L=T-V,$$

$$\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_k}\right)-\frac{\partial L}{\partial q_k}=0, \quad k=1, \ldots, n$$

## 物理代写|分析力学代写分析力学代考|拉格朗日方程的不变量

$$Q_k=G_k\left(q_1, \ldots, q_n, t\right), \quad k=1, \ldots, n$$
，反之，
$$q_k=g_k\left(Q_1, \ldots, Q_n, t\right), \quad k=1, \ldots, n .$$可逆性要求变换的雅可比矩阵满足以下条件:
$$\frac{\partial\left(q_1, \ldots, q_n\right)}{\partial\left(Q_1, \ldots, Q_n\right)} \equiv \operatorname{det}\left(\frac{\partial q_k}{\partial Q_l}\right)=\left(\frac{\partial\left(Q_1, \ldots, Q_n\right)}{\partial\left(q_1, \ldots, q_n\right)}\right)^{-1} \neq 0 .$$坐标变化(1.101)被称为点转换，因为它从配置空间映射点 $q s$ 的构型空间中的点 $Q$ 在数学术语中，一种双射可微的应用 $G$ 谁的逆矩阵 $g=G^{-1}$ 也是可微的被称为微分胚性，而 $Q \mathrm{~s}$ 它对构型空间是微分纯的 $q \mathrm{~s}$对Eq.(1.102)求时间导数，我们发现
$$\dot{q}k=\sum{l=1}^n \frac{\partial q_k}{\partial Q_l} \dot{Q}_l+\frac{\partial \dot{q}_k}{\partial t},$$

$$\frac{\partial \dot{q}_k}{\partial \dot{Q}_l}=\frac{\partial q_k}{\partial Q_l}$$

$$\bar{L}(Q, \dot{Q}, t)=L(q(Q, t), \dot{q}(Q, \dot{Q}, t), t)$$

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